\begin{answer}{rolltwodice}
The easiest thing to do here is to draw the matrix of possibilities.
(You don't actually have to fill in all the values like in the illustration; drawing a $6 \times 6$ block should be enough to figure out the answer.)
%
\begin{center}
\begin{tikzpicture}[scale=1.1]
% Draw the diagonal (must do this first
\foreach \x in {1,...,6}
{
    % Make the node a circle for space reasons, but don't draw it
    \filldraw[fill=lightgray] (\x,-\x) +(-.5,-.5) rectangle ++(.5,.5);
}
% Draw the matrix
\foreach \x in {1,...,6}
{
  % Make nodes on the outside of the matrix on the right side
  % Label them with o1, o2, etc.
  \foreach \y in {1,...,6}
  {
    \draw (\x,-\y) +(-.5,-.5) rectangle ++(.5,.5);
    % Make the node a circle for space reasons, but don't draw it
    \draw (\x,-\y) node[circle, inner sep=2mm] (n\y\x) {
                                           \drawdie[line hidden/.append style={fill=white}]{\y}
                                           \drawdie[line hidden/.append style={fill=white}]{\x}
                                           }; %reverse the x and y to give the expected behaviour
  }
}
\end{tikzpicture}
\end{center}
%
Now, depending on what the interviewer meant by the question, the answer is one of the following:
\begin{enumerate}
  \item
Asking for the probability of one die being larger than the other (and not caring which one) is analogous to asking for the probability that the two dice are \emph{not} the same.
That is everything but the diagonal,
\[
  \frac{36- 6}{36} =  \frac{30}{36} = \frac{5}{6}
  \text{.}
\]
  \item
If the interviewer wanted the probability of die $A$ being more than die $B$, then the possibilities are everything in the upper/lower triangle of the matrix (half of the answer given above),
\[
  \frac{15}{36} = \frac{5}{12}
  \text{.}
\]
\end{enumerate}
When you're given a question like this, your immediate reaction might be to start writing out the calculations---tedious sums or integrals.
Resist this urge.
This family of questions (quick probability calculations) can be easily solved by drawing a square on which you can calculate the area without doing integrals.
Other examples are the
\emph{Stick Breaking} problem (question \ref{q:stickbreak})
and the
\emph{Romeo and Juliet} problem (question \ref{q:romeojuliet}).

This simple example shows the importance of interview preparation.
You might think it a given that a rigorous solution is preferred to a back-of-the-envelope one, but you need to keep in mind that interviews aren't testing for job preparedness.
They are probably trying to test for several qualities, but they end up selecting the candidates who are the most prepared for brainteasers.
The format is akin to asking candidates to do a Rubik's cube and preferring the one who memorised the solution by rote and solves it in two minutes, rather than the one who takes the cube home and solves it over the course of three months without looking at the solution online.
Who would you rather have work for you?
Understand that brainteasers favour the former candidate and keep it in mind during preparation.
Use the pretty solution during the interview and leave the rigorous integrals for a Saturday night with a glass of whiskey in your study, while reflecting on the fact that people in quantitative finance are probably all brainteaser savants.
\end{answer}
